How much current is flowing in a wire 4.55 m long if the maximum force on it is 0.575 N when placed in a uniform 0.0700-T field?

1.81A

The magnitude of the magnetic force (F) acting on a current carrying wire is given by;

F = I x L x B sin θ      (i)

Where;

I = current on the wire

B = magnetic strength

L = length of the wire

θ = angle between the wire and the magnetic field

From the question, there is a maximum force (F = 0.575N) of on the wire. This means that the angle between the wire and the magnetic field is 90°.

Also;

the magnetic field(B) = 0.0700-T

the length(L) of the wire = 4.55m

Substitute these values into equation (i) as follows;

0.575 = I x 4.55 x 0.0700 x sin 90°

0.575 = I x 4.55 x 0.0700 x 1

0.575 = I x 0.3185

Solve for I;

I = 0.575 / 0.3185

I = 1.81A

Therefore, the current flowing in the wire is 1.81A

explanation:

newton's third law, for every action there is an equal and opposite reaction.

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